R E S E A R C H
Open Access
Factorizations for difference operators
Jeffrey Bergen
1, Mark Giesbrecht
2, Pappur N Shivakumar
3and Yang Zhang
3**Correspondence:
We consider the factorization problems of difference operators inC[x;
σ
] for an automorphismσ
of finite order. We study the factorization of regular polynomials in R[x] in the ring of such difference operators and obtain an analogue of thefundamental theorem of algebra for skew polynomial ring K[x;
σ
] over field K.Keywords: difference operators; factorizations
Linear differential operators (σ = idR, the identity map onR) and linear difference op-erators (δ= , the identically zero function) are special cases of above skew polynomials, which have been studied via algebraic methods since []. The study of skew polynomials is now an active and important area in algebra with many applications in other areas such as solving differential/difference equations, engineering, coding theory,etc.More recent re-sults and survey can be found in [–]. Factorization of skew polynomials is also a recent active area in computer algebra (see, for example, [–]). Algorithms for characteriza-tion of linear difference and differential equacharacteriza-tions are of great interest.
One of the great strengths of skew polynomials is their analogy with regular commuta-tive polynomials. A natural question thus arises: Is there an analogue of the fundamental theorem of algebra for skew polynomials? We provide a positive answer for difference op-erators with finite-order automorphisms over an algebraically closed field.
Throughout this paper,we assume thatR[x;σ]is a skew polynomial ring with automor-phismσas defined above.
2 Some basic results
It is clear that the polynomialsx– andx+ inR[x] factor uniquely into monic, linear
polynomials inC[x]. However, the situation is quite different when we look atR[x] as a subring of the skew polynomial ringC[x;σ]. Observe that ift∈Csuch that|t|= , and σ(t) is the complex conjugate oft, then
x– =x+σ(t)(x–t)
inC[x;σ]. Thusx– can be factored in an infinite number of ways inC[x;σ]. On the other
hand,x+ remains irreducible inC[x;σ]. In this paper we contrast how polynomials
factor in skew polynomial rings with how they factor in ordinary polynomial rings. We next introduce the terminology used throughout the remainder of this paper. We letKdenote a field andσan automorphism ofK. DefineKσ={a∈K|σ(a) =a}and, when
we examine the special case whereσ= , an important subset ofKσ is the set of norms
defined byN(K,σ) ={tσ(t)|t∈K}. When dealing with the fieldC, unless stated otherwise, we always assume thatσrepresents complex conjugation.
Forf(x) =anxn+· · ·+ax+a∈K[x;σ] andt∈K, the (standard) evaluation off(x) at
x=tis defined asf(t) =antn+· · ·+at+a. When we study polynomials over fields, one
of the first facts we prove is that linear factors correspond to roots. Note that this is not the case in skew polynomial rings: ifσdenotes complex conjugation, then
f(x) =x– = (x–i)(x–i)
inC[x;σ], yetf(i)= . To better understand the role of linear factors in skew polynomial rings, we defineσ-evaluationas follows.
Definition . Letf(x) =anxn+an–xn–+· · ·+ax+ax+a∈K[x;σ] andt∈K. Then we
define theσ-evaluation f(t;σ)∈Kasf(t;σ) =an(tσ(t)· · ·σn–(t)) +an–(tσ(t)· · ·σn–(t)) +
· · ·+a(tσ(t)) +at+a. Iff(t;σ) = , we say thattis aσ-rootoff(x).
Observe that ifσ= idR, andK[x;σ] =K[x] is an ordinary polynomial ring, then aσ-root is the same as an ordinary root. Lemma . and Theorem . show us that the role of σ-roots in the skew polynomial case generalizes the role of roots in the ordinary case.
Lemma .([]) If f(x) =anxn+· · ·+ax+a∈K[x;σ]and t∈K,then there exists a
unique q(x)∈K[x;σ]such that f(x) =q(x)(x–t) +f(t;σ).
Whenf(x) =q(x)(x–t), instead of simply saying thatx–tis a factor off(x), we may sometimes use the more precise terminology thatx–tis a (right) factor off(x). We can now see the relationship between (right) factors andσ-roots.
Theorem . Suppose f(x)∈K[x;σ].Then x–t is a right factor of f(x)if and only if t is a
σ-root of f(x).
Proof Using Lemma ., there exists a uniqueq(x)∈K[x;σ] such thatf(x) =q(x)(x–t) +
a (right) factor off(x). Conversely, ifx–tis a (right) factor off(x), thenf(x) =h(x)(x–t) for someh(x)∈K[x;σ]. But then, by Lemma ., the quotienth(x) and remainder () are
unique, sof(t;σ) = .
For ordinary polynomials of degree , being reducible is equivalent to having a root. In our situation, we now show that it is equivalent to having aσ-root.
Corollary . Suppose f(x)∈K[x;σ]has degree;then f(x)is reducible inK[x;σ]if and only if f(x)has aσ-root inK.
Proof One direction follows immediately from Theorem ., for iff(x) has aσ-root inK, then it has a right factor of degree one inK[x;σ] and is therefore reducible. In the other direction, iff(x) is reducible and has degree , then it is easy to see that it must have a monic, linear right factor. Theorem . now asserts thatf(x) has aσ-root inK.
One of the examples that motivated this paper was the fact thatx+ remains irre-ducible inC[x;σ]. Thus the situation of quadratic polynomials whereσ= is of
particu-lar interest. We can now determine precisely that irreducible quadratics inKσ[x] remain
irreducible inK[x;σ].
Corollary . Let f(x) =x+ax+b∈Kσ[x]and supposeσ= .Then f(x)is reducible in
K[x;σ]if and only if either(i)f(x)is reducible inKσ[x],or(ii)a= and–b∈N(K,σ).
Proof Iff(x) is reducible inKσ[x], then it is certainly reducible inK[x;σ]. Furthermore, if
a= and –b∈N(K,σ), then –b=tσ(t) for somet∈Kand
f(x) =x– (–b) =x–tσ(t)=x+σ(t)(x–t).
This proves one half of the result.
To see the converse, Corollary . tells us that iff(x) is reducible inK[x;σ], then it has a σ-roott∈K. Iftis aσ-root, then it satisfies the equation
tσ(t) +at+b= .
Sinceσ= , we know thattσ(t)∈Kσ. However,b∈Kσ, therefore it follows thatat∈Kσ.
Ifa= , then the fact thata∈Kσ implies thatt∈Kσ. Sincetis an ordinary root off(x) in
Kσ, we see thatf(x) is reducible inKσ[x]. On the other hand, supposea= . Our equation
above now tells us that –b=tσ∈N(K,σ), thereby concluding the proof. The next corollary completely describes the situation for the examplesx– ,x+ ∈
R[x]⊆C[x;σ] that were mentioned at the start of this paper.
Corollary . Let f(x) =x+ax+b∈R[x].
(i) f(x)is reducible inC[x;σ]if and only iff(x)is reducible inR[x].
(ii) Iff(x)is reducible,then the factorization off(x)inC[x;σ]into monic,linear factors is unique whena= ,whereasf(x)factors an infinite number of ways into monic,
Proof In light of Corollary ., in order to prove part (i), it suffices to show that ifa= and –b∈N(C,σ), thenf(x) is reducible inR[x]. However, in this situation,N(C,σ) is equal to the set of positive real numbers. Therefore, if –b∈N(C,σ), then –b=cfor somec∈R.
Thus
f(x) =x– (–b) = (x+c)(x–c),
as required.
For the proof of part (ii), letcis(θ) =cosθ+isinθfor everyθ∈R. Then, when we are in the case wherea= , we can rewrite the equation above as
f(x) =x– (–b) = (x+c)(x–c) =x+ccis(–θ)x–ccis(θ).
By lettingθrange between and π, we can see thatf(x) does factor an infinite number of ways inC[x;σ].
Finally, suppose a= ; the argument in the proof of Corollary . indicates that any factorization off(x) actually takes place inR[x]. Since there is only one way to factorf(x) inR[x], it follows that the factorization off(x) inC[x;σ] is also unique.
When we look back at Corollaries . and ., it becomes natural to look for an example where reducibility inK[x;σ] is not equivalent to reducibility inKσ[x].
Example . LetK=Q(i) and once again letσ denote complex conjugation. In this case,
x– andx– are irreducible inKσ[x] =Q[x]. However, since , ∈N(Q,σ), these
poly-nomials become reducible inQ(i)[x;σ] as we have
x– =x+ ( –i)x– ( +i) and x– =x+ ( –i)x– ( +i).
On the other hand, since –, /∈N(Q(i),σ), the polynomialsx+ andx– are irre-ducible inQ[x] and remain irreducible inQ(i)[x;σ].
3 Main theorems
In order to complete our description of the factoring of polynomials inC[x;σ], we first need to prove a result that holds in any algebraically closed field.
Theorem . LetKbe an algebraically closed field and letσbe an automorphism ofKof order n.Then every non-constant reducible skew polynomial inK[x;σ]can be written as a product of irreducible skew polynomials of degree less than or equal to n.
Proof Sinceσhas ordern,Khas dimensionnover the subfieldKσ. It is well known that the
centre ofK[x;σ] is the polynomial ringKσ[xn]. Now suppose thatf(x) is a non-constant
element ofK[x;σ]. SinceK[x;σ] is a finite module over the Noetherian ringKσ[xn],K[x;σ]
contains no infinite direct sums ofKσ[xn]-submodules. As a result, the sum
f(x)Kσxn+f(x)Kσxn+f(x)Kσxn+· · ·
cannot be direct. Therefore, there exista(x), . . . ,am(x)∈Kσ[xn], with at least two nonzero
ai(x), such that
By cancelling any common factors off(x) on the left, we obtain
aj(x) +f(x)aj+(x) +· · ·+f(x)m–jam(x) = .
Observe thataj(x) is a nonzero element ofKσ[xn]∩f(x)K[x;σ]. Therefore, there exists
g(x)∈K[x;σ] such thatf(x)g(x) is a nonzero element ofKσ[xn].
SinceKis algebraically closed andn-dimensional overKσ, non-constant polynomials in
Kσ[y] can all be factored into a product of polynomials of degree at mostn. If we lety=xn,
then
f(x)g(x)∈Kσxn=Kσ[y].
Therefore, there exist non-constanth(y), . . . ,hm(y)∈Kσ[y], all of degree at mostn, such
that
f(x)g(x) =h(y)· · ·hm(y).
For eachi, the polynomialhi(y) =hi(xn)∈Kσ[x] can also be factored inKσ[x] as a product
of non-constant polynomials of degree at mostn. As a result, we have actually factored
f(x)g(x) inK[x,σ] as a product of polynomials of degree at mostn.
By Theorem of Chapter II in [], the degrees of factors off(x)g(x) are at mostn. Thusf(x) has been factored inK[x;σ] as a product of polynomials of degree at mostn.
Using Theorem . and Corollary ., we now have the following.
Corollary . Every non-constant polynomial inC[x;σ]is a product of linear and irre-ducible polynomials.Furthermore,the quadratic f(x) =x+ax+ais irreducible inC[x;σ] if and only if there does not exist some t∈Csuch that tσ(t) +at+a= .
Proof Theorem . asserts that every non-constant polynomial inC[x;σ] is a product of linear and irreducible polynomials. In addition, Corollary . tells us thatf(x) =x+ax+ ais irreducible inC[x;σ] if and only iff(x) does not have aσ-root inC. However, when we look at the definition ofσ-roots in the quadratic case, it is clear thatf(x) does not have aσ-root if and only if there does not existt∈Csuch thattσ(t) +at+a= .
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Author details
1Department of Mathematical Sciences, DePaul University, Chicago, IL 60614, USA.2David R. Cheriton School of
Computer Science, University of Waterloo, Waterloo, ON, Canada.3Department of Mathematics, University of Manitoba,
Winnipeg, MB, Canada.
Acknowledgements
We would like to thank the anonymous referees for their careful reading and valuable comments that have improved the readability of this article. The research of the first author was supported by a grant from the University Research Council at DePaul University.
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